Hint

Let $P,Q$ be the feet of perpendiculars from $D$ onto $AB,AC.$ Let $BD=x,CD=3C,AD=d,\angle DAB=\alpha,\angle DAC=\beta.$ Then, using $\textcolor{red}{\textit{Ratio Lemma}}$ & $\textcolor{red}{\textit{Stewart's Theorem,}}$ it's easy to show that $$\frac{\sin\beta}{\sin\alpha}=3,d=\sqrt{13}x.$$Now, we have $$\frac{[BDE]}{[CDF]}=\frac{DP\cdot BE}{DQ\cdot CF}=\frac{\sin\alpha\cdot BE}{\sin\beta\cdot CF}=\frac 13\frac{BE}{CF}.$$Now, using $\textcolor{red}{\textit{Law of Sines,}}$ we can show that $$BE=\frac{d\sin 2\beta}{\sqrt 3\cos\alpha},CF=\frac{d\sin 2\alpha}{\sqrt 3\cos \beta}.$$Therefore, $$\frac{BE}{CF}=\frac{\sin 2\beta}{\cos\alpha}\cdot\frac{\cos \beta}{\sin 2\beta}=\frac{\sin \beta\cos^2\beta}{\sin\alpha\cos^2\alpha}=\frac{3\cos^2\beta}{\cos^2\alpha}.$$But $$\sin\beta=\sin(60^{\circ}-\alpha)=\frac{\sqrt 3}{2}\cos\alpha-\frac 12\sin\alpha=3\sin\alpha$$$$\implies \cot\alpha=\frac{7}{\sqrt 3}.$$Thus, $$\frac{\cos^2\beta}{\cos^2\alpha}=\frac{1-\sin^2\beta}{1-\sin^2\alpha}=\frac{\cot^2\alpha-8}{\cot^2\alpha}=\frac{25}{49}.$$Hence, $$\boxed{\frac{[BDE]}{[CDF]}=\frac 13\cdot 3\cdot \frac{25}{49}=\frac{25}{49}.}$$$\textcolor{green}{\cal{QED.}}$

Suppose that the sidelength of the triangle is $4$ units. Then $BD = 1$, $CD = 3$. Let $BE = x$ and $CF = y$. Then $DE = AE = 4-x$, and $DF = AF = 4-y$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7.809803807407429cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -22.9138328600633, xmax = 81.21688457203577, ymin = -14.608910794874358, ymax = 32.651176945478994; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((12.012926477213199,7.554339014267875)--(14.039796202813942,6.998299855703401)--(14.595835361378416,9.025169581304144)--(12.568965635777673,9.581208739868618)--cycle, linewidth(0.4)); /* draw figures */ draw((8.380582831437453,10.730224298616967)--(7.925274540752694,-7.345938837016567), linewidth(0.8) + wrwrwr); draw((7.925274540752694,-7.345938837016567)--(30.1601734140066,4.75534339312557), linewidth(0.8) + wrwrwr); draw((17.212656730802657,26.50835631675381)--(7.925274540752694,-7.345938837016567), linewidth(0.8) + wrwrwr); draw((8.380582831437453,10.730224298616967)--(30.1601734140066,4.75534339312557), linewidth(0.8) + wrwrwr); draw((-1.8102716678300883,-7.475312645387078)--(8.380582831437453,10.730224298616967), linewidth(0.8) + wrwrwr); draw((8.380582831437453,10.730224298616967)--(17.212656730802657,26.50835631675381), linewidth(0.8) + wrwrwr); draw((17.212656730802657,26.50835631675381)--(30.1601734140066,4.75534339312557), linewidth(0.8) + wrwrwr); draw((37.13191316650104,-6.957817411905031)--(30.1601734140066,4.75534339312557), linewidth(0.8) + wrwrwr); draw((-1.8102716678300883,-7.475312645387078)--(7.925274540752694,-7.345938837016567), linewidth(0.8) + wrwrwr); draw((7.925274540752694,-7.345938837016567)--(37.13191316650104,-6.957817411905031), linewidth(0.8) + wrwrwr); /* dots and labels */ dot((17.212656730802657,26.50835631675381),linewidth(2.pt) + dotstyle); label("$A$", (17.608967739107502,26.904667325058675), NE * labelscalefactor); dot((-1.8102716678300883,-7.475312645387078),linewidth(2.pt) + dotstyle); label("$B$", (-3.890904461430554,-10.249489703521009), NE * labelscalefactor); dot((37.13191316650104,-6.957817411905031),linewidth(2.pt) + dotstyle); label("$C$", (37.523595906426166,-6.583612876701147), NE * labelscalefactor); dot((7.925274540752694,-7.345938837016567),linewidth(2.pt) + dotstyle); label("$D$", (8.097503539791127,-10.645800711825858), NE * labelscalefactor); dot((8.380582831437453,10.730224298616967),linewidth(2.pt) + dotstyle); label("$E$", (4.530704465047486,8.872516447188001), NE * labelscalefactor); dot((30.1601734140066,4.75534339312557),linewidth(2.pt) + dotstyle); label("$F$", (30.588153261091307,5.107561868291926), NE * labelscalefactor); label("4-x", (5.422404233733396,0.15367426448130206), NE * labelscalefactor,wrwrwr); label("4-y", (19.78867828478417,-2.6205027936526477), NE * labelscalefactor,wrwrwr); label("x", (1.6574496548373312,2.630618066386614), NE * labelscalefactor,wrwrwr); label("4-x", (10.47536958962022,19.473835919342736), NE * labelscalefactor,wrwrwr); label("4-y", (24.049021624061297,17.095969869513638), NE * labelscalefactor,wrwrwr); label("y", (34.74941884829222,-0.3417144958997604), NE * labelscalefactor,wrwrwr); label("1", (2.6482271755994535,-9.754100943139946), NE * labelscalefactor,wrwrwr); label("3", (23.058244103299174,-9.655023191063734), NE * labelscalefactor,wrwrwr); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We have that $AEDF$ is a kite, so $\angle EDF = \angle EAF = 60^\circ$. Thus, $\angle BDE = 120^\circ - \angle CDF = \angle CFD$. Hence, we have that $\triangle BED \sim \triangle CDF$ (by AA criterion). Thus we have $$\frac{x}{3} = \frac{4-x}{4-y} = \frac{1}{y}.$$Solving this gives us $\displaystyle x = \frac{15}{7}, y = \frac{7}{5}.$ Thus, since the ratio of areas of similar triangles is the square of the corresponding sides, we have $$ \frac{[BDE]}{[CDF]} = \left(\frac{x}{3} \right)^2 = \frac{25}{49},$$as desired. $\blacksquare$