Answer
$$\frac{\partial}{\partial x}h(x,y,z,t)=2xy\cos\frac{z}{t}.$$
$$\frac{\partial}{\partial y}h(x,y,z,t)=x^2\cos(z/t).$$
$$\frac{\partial}{\partial z}h(x,y,z,t)=-\frac{x^2y}{t}\sin\frac{z}{t}.$$
$$\frac{\partial}{\partial t}h(x,y,z,t)=\frac{x^2yz}{t^2}\sin\frac{z}{t}.$$
Work Step by Step
Regard $y,z$ and $t$ as constant and differentiate with respect to $x$:
$$\frac{\partial}{\partial x}h(x,y,z,t)=\frac{\partial}{\partial x}(x^2y\cos(z/t))=y\cos(z/t)\frac{\partial}{\partial x}x^2=y\cos(z/t)\cdot 2x=2xy\cos\frac{z}{t}.$$
Regard $x,z$ and $t$ as constant and differentiate with respect to $y$:
$$\frac{\partial}{\partial y}h(x,y,z,t)=\frac{\partial}{\partial y}(x^2y\cos(z/t))=x^2\cos(z/t)\frac{\partial}{\partial y}y=x^2\cos(z/t).$$
Regard $x,y$ and $t$ as constant and differentiate with respect to $z$:
$$\frac{\partial}{\partial z}h(x,y,z,t)=\frac{\partial}{\partial z}(x^2y\cos(z/t))=x^2y\frac{\partial}{\partial z}\cos(z/t)=x^2y(-\sin(z/t))\frac{\partial}{\partial z}\frac{z}{t}=-x^2y\sin(z/t)\frac{1}{t}\frac{\partial}{\partial z}z=-\frac{x^2y}{t}\sin\frac{z}{t}.$$
Regard $x,y$ and $z$ as constant and differentiate with respect to $t$:
$$\frac{\partial}{\partial t}h(x,y,z,t)=\frac{\partial}{\partial t}(x^2y\cos(z/t))=x^2y\frac{\partial}{\partial t}\cos(z/t)=x^2y(-\sin(z/t))\frac{\partial}{\partial t}\frac{z}{t}=-x^2y\sin(z/t)z\frac{\partial}{\partial t}\frac{1}{t}=-x^2yz\sin(z/t)\left(-\frac{1}{t^2}\right)=\frac{x^2yz}{t^2}\sin\frac{z}{t}.$$
